Newton's laws applications — elevator, pulley, spring scale, connected bodies

Question

A person of mass 60 kg stands on a weighing scale inside an elevator. What does the scale read when the elevator: (a) accelerates upward at 2 m/s², (b) accelerates downward at 2 m/s², (c) moves at constant velocity, (d) is in free fall? Take $g = 10$ m/s².

(CBSE Class 11 / NEET / JEE Main pattern)

Solution — Step by Step

A weighing scale measures the normal force (or apparent weight), not the actual gravitational weight. The reading depends on the acceleration of the elevator.

flowchart TD
    A["Person in Elevator"] --> B{"Elevator\nacceleration?"}
    B -->|"a upward"| C["Apparent weight\n= m(g+a)\nFeels heavier"]
    B -->|"a downward"| D["Apparent weight\n= m(g-a)\nFeels lighter"]
    B -->|"a = 0\n(constant v or rest)"| E["Apparent weight\n= mg\nNormal"]
    B -->|"a = g downward\n(free fall)"| F["Apparent weight\n= 0\nWeightlessness"]

Taking upward as positive, for the person:

N - mg = ma

N = m(g + a)

where $a$ is positive for upward acceleration and negative for downward.

(a) Accelerating upward ( $a = +2$ m/s²):

N = 60(10 + 2) = \mathbf{720 \text{ N}} \quad (\text{scale reads } 72 \text{ kg})

(b) Accelerating downward ( $a = -2$ m/s²):

N = 60(10 - 2) = \mathbf{480 \text{ N}} \quad (\text{scale reads } 48 \text{ kg})

(c) Constant velocity ( $a = 0$ ):

N = 60 \times 10 = \mathbf{600 \text{ N}} \quad (\text{scale reads } 60 \text{ kg})

(d) Free fall ( $a = -g = -10$ m/s²):

N = 60(10 - 10) = \mathbf{0 \text{ N}} \quad (\text{weightlessness!})

Why This Works

The scale measures the contact force between the person and the scale surface, which is the normal force $N$ . In an accelerating frame, $N \neq mg$ because a net force is needed to accelerate the person along with the elevator. When the elevator accelerates up, the floor must push harder (more than $mg$ ) to accelerate the person upward. When it accelerates down, less force is needed.

In free fall, both the person and the elevator accelerate at $g$ downward — there is no need for any contact force, so $N = 0$ . This is the origin of weightlessness in space (astronauts are in continuous free fall around the Earth).

Alternative Method — Working in the Elevator Frame (Pseudo Force)

In the non-inertial elevator frame, add a pseudo force $= -ma$ on the person:

Apparent weight $= mg + m(-a) = m(g-a)$ for downward acceleration $a$ .

This gives the same result but uses the concept of pseudo (fictitious) forces, which is useful for more complex problems involving non-inertial frames.

NEET and JEE Main love this topic. The formula $N = m(g \pm a)$ covers all elevator problems. Use $+a$ when the elevator accelerates upward (or decelerates while going down), and $-a$ when it accelerates downward (or decelerates while going up). The direction of acceleration matters, not the direction of velocity.

Common Mistake

Students confuse the direction of velocity with the direction of acceleration. An elevator moving upward but decelerating has a downward acceleration — the apparent weight decreases, not increases. Similarly, an elevator moving downward but slowing down has an upward acceleration — apparent weight increases. Always determine the acceleration direction first, then apply the formula.

Question

Solution — Step by Step

Why This Works

Alternative Method — Working in the Elevator Frame (Pseudo Force)

Common Mistake

Try These Next